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1991 AIME Problems/Problem 2

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Problem

Rectangle ABCD_{}^{} has sides \overline {AB} of length 4 and \overline {CB} of length 3. Divide \overline {AB} into 168 congruent segments with points A_{}^{}=P_0, P_1, \ldots, P_{168}=B, and divide \overline {CB} into 168 congruent segments with points C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B. For 1_{}^{} \le k \le 167, draw the segments \overline {P_kQ_k}. Repeat this construction on the sides \overline {AD} and \overline {CD}, and then draw the diagonal \overline {AC}. Find the sum of the lengths of the 335 parallel segments drawn.

Solution

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The length of the diagonal is \sqrt{3^2 + 4^2} = 5 (a 3-4-5 right triangle). For each k, \overline{P_kQ_k} is the hypotenuse of a 3-4-5 right triangle with sides of 3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}. Thus, its length is 5 \cdot \frac{k}{168}.

The sum we are looking for is 2 \cdot \left(\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}\right) + 5 = \frac{5}{84}\sum_{k=1}^{167}k + 5. Using the formula for the sum of the first natural numbers, we find that the solution is \frac{5}{84} \cdot \frac{168 \cdot 167}{2} + 5 = 5 \cdot 167 + 5 = \boxed{840}.

See also

1991 AIME (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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